# Full text of "History Of The Theory Of Numbers - I"

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```CHAP, xivi                      METHODS OF FACTORING.                               365
A. Cunningham67 and J. Cullen listed the 188 prime numbers x*+lS48y* between 107 and 101-105, with a; prime to lS4&y.
A. Cunningham68 noted that two representations of N by vxk+pyk lead to factors of N under certain conditions.
A. Cunningham69 recalled that an idoneal number I has the property that, if an odd number N is expressible in only one way hi the form N = mx2+ny2j where mn = 7, and mx2 is prime to ny2, then N is a prime or the square of a prime. Euler's largest I is 1848. There is no larger I under 50000, a computation checked by J. Cullen. Cunningham noted on the proof-sheets of this history that this limit has been extended to 100 000.
A. Cunningham70 noted conditions that an odd prime be expressible by t2^qu2 when q or — q is idoneal.
F. N. Cole71 discussed Seelhoff's56 method of factoring.
Al. Laparewicz72 described and applied Gauss' 48>49 two methods.
P. Meyer73 discussed Euler's theorem that, if n is idoneal, a number representable only once by x2+ny2 is a prime.
R. Burgwedel74 gave an exposition and completion of the method of Euler37"43 and an exposition of the methods of Gauss.48'49
L. Valroff stated and A. Cunningham740 proved that CDz2-a2)(Di/2--a2) = I>22 — a2 implies that one factor is composite unless z2 = 2/2 = 4 when a = I j D = 2, and in the remaining cases if the two factors are distinct and > 1.
A. Ge*rardin75 gave a method illustrated for N = a2- 5-292, where a = 6326. We shall have a second such representation N= (a-f-5x)2 — 5?/2 if
Use is made of various moduli m = 4, 3, 7, 25, .... On square-ruled paper, mark x = 0, 1, 2, ... at the head of the columns. On the line for modulus m, shade the square under the heading x when x makes E a quadratic non-residue of m. Then examine the column in which occurs no shaded square. Up to z^l5, these are x = 0 (excluded), and x = 4, which gives N = 63462 — 5-22T2 and the factor 992 — 5-22. The same diagram serves for all numbers 1050 #+671, our N being given by #=38108. To apply the method to N = (2z)4+l = (4z2-|-l)2 — 2(2x)2, seek a second representation N=(4x2 -f2p+l)2--2(2u)2. The condition is (2p+l)x2+|p(p+l)='U2, solutions of which are found for p = 1, 8, 9, . . . , 62, 352, ... Or we may choose x, say x = 48, and find p = 8, w = 198.
67Brit. Assoc. Reports, 1901, 552.    The entry 10098201 is erroneous.
«8Proe. London Math. Soc., 33, 1900-1, 361.